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Theorem 11. Let \( D \) be strictly convex. For all \( 0 < r < R,{D}_{r} \) is strictly convex. | Proof. Suppose that \( \partial {D}_{r} \) contains a segment \( l \) . Let \( p \) be a point in the interior of \( l \) . As it is \( r \) -accessible, there is a segment \( \left\lbrack {P, Q}\right\rbrack \) of length \( r \) through \( p \), where \( P, Q \in \partial D \) . By Lemma \( 5,{v}_{P},{v}_{Q} \) are no... | Yes |
Proposition 12. If \( {I}_{D} \) has non-empty interior, then \( \partial D \) contains two open subsets which are included in parallel hyperplanes, which are at distance \( R \) . | Proof. Consider an interior point \( p \in {I}_{D} \), so \( \mathbf{r}\left( p\right) = R \) . Take a segment \( l = \left\lbrack {P, Q}\right\rbrack \) of length \( R \) with endpoints \( P, Q \in \partial D \) . Let \( {v}_{P},{v}_{Q} \) be vectors orthogonal to the supporting hyperplanes at \( P, Q \) . By Lemma 5,... | Yes |
Theorem 13. Let \( D \) be a strictly convex bounded domain, \( R = \max \mathbf{r} \). Then \( {I}_{D} \) is a point or a segment. | Proof. Suppose that \( {I}_{D} \) is not a point. As it is convex by Theorem 6, it contains a maximal segment \( \sigma \). Let us see that it cannot contain two different (intersecting) segments. Let \( p \in \sigma \) be an interior point of the segment. By Lemma 5, if we draw the segment \( \left\lbrack {P, Q}\right... | Yes |
Corollary 14. Suppose \( D \) is a planar convex bounded domain (not necessarily strictly convex). If \( {I}_{D} \) is not a point then there are two non-corner points \( P, Q \in \) \( \partial D \) with parallel tangent hyperplanes which are moreover perpendicular to \( \overrightarrow{PQ} \) . | Proof. Following the proof of Theorem 13, we only have to rule out case (2). As the hyperplane \( {H}_{v} \) is now of dimension 1, we have \( \sigma \subset \left\lbrack {P, Q}\right\rbrack = {H}_{v} \cap \bar{D} \) . But Lemma 5 says also that \( {E}_{R} \cap \left\lbrack {P, Q}\right\rbrack = \{ p\} \) . So \( {E}_{... | Yes |
Lemma 15. Fix \( \lambda \in \mathbb{R} \) . Let \( D \) be the domain with boundary the half-lines \( \left( {x,0}\right) \) , \( x \geq 0 \) and \( \left( {{\lambda y}, y}\right), y \geq 0 \) . Let \( r > 0 \) . Then the boundary of \( {D}_{r} \) is the curve:\n\n\[ \left\{ \begin{array}{l} x = r\left( {{\cos }^{3}\t... | Proof. \( D \) is not a bounded domain, but the theory works as well in this case. To find the boundary of \( {D}_{r} \), we need to take the envelope of the segments of length \( r \) with endpoints laying on the half-rays, according to Proposition 7. Two points at \( \left( {a,0}\right) \) and \( \left( {{\lambda b},... | Yes |
Let \( D \) be a planar convex polygon. Then the sets \( {D}_{r},0 < r < R \) , and \( {I}_{D} \) if it is not a point, have boundaries which are piecewise \( {C}^{1} \), and whose pieces are \( \lambda \) -bows and (possibly) segments in the sides of \( \partial D \) . In particular, these domains are strictly convex ... | Proof. Let \( {l}_{1},\ldots ,{l}_{k} \) be the lines determined by prolonging the sides of the polygon. Consider \( {l}_{i},{l}_{j} \) . If they intersect, consider the sector that they determine in which \( D \) is contained. Lemma 15 provides us with a (convex) region \( {D}_{r}^{ij} \) . If \( {l}_{i},{l}_{j} \) ar... | Yes |
Lemma 1. The perpendicular bisector segment through the midpoint of one side terminates at a point on the longer of the remaining two sides (or at their intersection if these sides are equal). | Proof. Let the perpendicular bisector pass through the midpoint \( F \) of the side \( {AB} \) of triangle \( {ABC} \) . Consider the case when both angles \( A \) and \( B \) are acute. In Figure \( 1,{AF} = {FB} \) and \( {FE} \geq {FD} \) . We show that \( {BC} \geq {AC} \) . Now,\n\n\[ \n\tan A = \frac{EF}{AF} \geq... | Yes |
Theorem 2. Let \( a \geq b \geq c \) . \[ {p}_{a} = \frac{2a\Delta }{{a}^{2} + {b}^{2} - {c}^{2}},\;{p}_{b} = \frac{2b\Delta }{{a}^{2} + {b}^{2} - {c}^{2}},\;{p}_{c} = \frac{2c\Delta }{{a}^{2} - {b}^{2} + {c}^{2}}. \] | Proof. (i) Figure 3 illustrates the case where \( \angle A \) is acute; the proof is identical when \( \angle A \) is right or obtuse. We have \( \tan C = \frac{{p}_{a}}{\frac{a}{2}} \) since by Lemma 1 bisector \( {p}_{a} \) meets side \( b \) because \( b \geq c \) . We know from [2] (which applies since \( \angle C ... | Yes |
Theorem 4. Let \( a \geq b \geq c \) .\n\n(i) \( {p}_{a} \geq {p}_{b} \) ;\n\n(ii) \( {p}_{c} \geq {p}_{b} \) ;\n\n(iii) \( {p}_{a} \geq {p}_{c},{p}_{a} = {p}_{c} \) and \( {p}_{a} < {p}_{c} \) are all possible. | Proof. (i) By Theorem \( 2,\frac{{p}_{a}}{{p}_{b}} = \frac{a}{b} = \frac{\sin A}{\sin B} \) . Since \( A \geq B \) and \( A + B < \pi \) , \( \sin A \geq \sin B \) . It follows that \( {p}_{a} \geq {p}_{b} \) .\n\n(ii) From Figures 4 and 5,\n\n\[ \frac{{p}_{b}}{{p}_{c}} = \frac{\frac{b}{2}\tan C}{\frac{c}{2}\tan B} = \... | Yes |
Theorem 7. Given positive \( {p}_{a},{p}_{b},{p}_{c} \) satisfying \( {p}_{a} \geq {p}_{b} \) and \( {p}_{c} \geq {p}_{b} \), there are no more than two non-congruent triangles with \( a \geq b \geq c \) and perpendicular bisector segments of lengths \( {p}_{a},{p}_{b},{p}_{c} \) . | Proof. By Theorem 2, \( \frac{{p}_{a}}{{p}_{b}} = \frac{a}{b} \), and\n\n\[ \frac{{p}_{c}}{{p}_{b}} = \frac{\frac{c}{b} \cdot \left( {{a}^{2} + {b}^{2} - {c}^{2}}\right) }{{a}^{2} - {b}^{2} + {c}^{2}} = \frac{\frac{c}{b}\left( {{\left( \frac{a}{b}\right) }^{2} + 1 - {\left( \frac{c}{b}\right) }^{2}}\right) }{{\left( \f... | Yes |
Theorem 8. Given \( {p}_{a} > {p}_{b} \) and \( {p}_{c} > {p}_{b} \), let \( \alpha = \frac{{p}_{a}}{{p}_{b}} \) and \( \gamma = \frac{{p}_{c}}{{p}_{b}} \) . The number of triangles with perpendicular bisector segments of lengths \( {p}_{a},{p}_{b},{p}_{c} \) is\n\n2 if and only if \( \gamma \in \left( {\frac{{\alpha }... | Proof. Note that \( f\left( {\alpha - 1}\right) = {2\alpha }\left( {\alpha - 1}\right) \left( {\gamma - 1}\right) > 0 \) and \( f\left( 1\right) = {\alpha }^{2}\left( {\gamma - 1}\right) > 0 \) .\n\nIf the local minimum \( {x}_{0} \in \left( {\alpha - 1,1}\right) \), then the number of roots of \( f\left( x\right) \) i... | Yes |
Corollary 9. There are a maximum of four triangles with the three segments of given lengths as perpendicular bisector segments. | We conclude this paper by giving explicit examples showing that the number of triangles in Corollary 9 can be any of \( 0,1,2,3,4 \) .\n\n(i) \( n = 4 \) : Consider \( \left( {{p}_{1},{p}_{2},{p}_{3}}\right) = \left( {{20},{18},{15}}\right) \).\n\nIf \( \left( {{p}_{a},{p}_{b},{p}_{c}}\right) = \left( {{20},{15},{18}}\... | Yes |
Proposition 1. The polynomial \( {L}_{a}{L}_{b}{L}_{c} - L\left( {P}_{1}\right) L\left( {P}_{2}\right) L\left( {P}_{3}\right) {xyz} \) is divisible by \( {px} + {qy} + {rz} \) . | Proof. If we put \( x = - \frac{{qy} + {rz}}{p} \), then\n\n\[ \n{L}_{a} = \left( {q{g}_{1} + r{h}_{1}}\right) x - {f}_{1}\left( {{qy} + {rz}}\right) = \left( {q{g}_{1} + r{h}_{1}}\right) x + {f}_{1} \cdot {px} = L\left( {P}_{1}\right) x, \n\] \n\n\[ \n{L}_{b} = L\left( {P}_{2}\right) y \n\] \n\n\[ \n{L}_{c} = L\left( ... | Yes |
Proposition 2. The conic defined by the symmetric matrix \( M \)\n\n(a) degenerates into a pair of lines if \( \det M = 0 \) ,\n\n(b) is a parabola if \( \det M \neq 0 \) and \( \chi \left( M\right) = 0 \) ,\n\n(c) has center \( Q = G{M}^{\# } \) if \( \left( {\det M}\right) \chi \left( M\right) \neq 0 \) . | Proof. Let \( Q = G{M}^{\# } \) . For arbitrary \( 1 \times 3 \) matrix \( P \) and real number \( t \) ,\n\n\[ \left( {Q + {tP}}\right) M{\left( Q + tP\right) }^{\mathrm{t}} = {QM}{Q}^{\mathrm{t}} + \left( {{PM}{Q}^{\mathrm{t}} + {QM}{P}^{\mathrm{t}}}\right) t + \left( {{PM}{P}^{\mathrm{t}}}\right) {t}^{2} \]\n\n\[ = ... | Yes |
Lemma 4. Let \( \mathcal{L} \) be the line \( {px} + {qy} + {rz} = 0 \) . The matrix \( M\left( \mathcal{L}\right) \) has (a) determinant \( \det M\left( \mathcal{L}\right) = - \frac{1}{4}{\left( p + q + r\right) }^{4}\left( {{qr} + {rp} + {pq}}\right) \) , (b) adjoint matrix | \[ M{\left( \mathcal{L}\right) }^{\# } = \frac{{\left( p + q + r\right) }^{2}}{4}\left( \begin{matrix} - {p}^{2} & {2qr} + {2rp} + {pq} & {2qr} + {rp} + {2pq} \\ {2qr} + {2rp} + {pq} & - {q}^{2} & {qr} + {2rp} + {2pq} \\ {2qr} + {rp} + {2pq} & {qr} + {2rp} + {2pq} & - {r}^{2} \end{matrix}\right) \] and (c) characterist... | Yes |
Proposition 6. If \( \mathcal{L} : {px} + {qy} + {rz} = 0 \) is the tangent to \( {\mathcal{E}}_{0}^{ * } \) at \( P \), then the parabola \( \mathcal{Q}\left( {\mathcal{L}, G}\right) \) and the ellipse \( {\mathcal{E}}_{0}^{ * } \) have\n\n(i) a common tangent parallel to \( \mathcal{L} \) at the antipode of \( P \) o... | Proof. We take the matrices of the dual ellipses \( {\mathcal{E}}_{0} \) and \( {\mathcal{E}}_{0}^{ * } \) to be\n\n\[ M = \left( \begin{matrix} - 1 & 5 & 5 \\ 5 & - 1 & 5 \\ 5 & 5 & - 1 \end{matrix}\right) \;\text{ and }\;{M}^{ * } = \left( \begin{matrix} - 4 & 5 & 5 \\ 5 & - 4 & 5 \\ 5 & 5 & - 4 \end{matrix}\right) \... | Yes |
Proposition 8. The conic \( {\mathcal{E}}_{1} \) is an ellipse with center \( G \) . Its dual conic is a circle also with center \( G \) . | Proof. The matrix of the conic \( {\mathcal{E}}_{1} \) being\n\n\[ \n{M}_{1} = \left( \begin{matrix} {S}_{A} & {S}_{A} + {S}_{B} + 2{S}_{C} & {S}_{A} + 2{S}_{B} + {S}_{C} \\ {S}_{A} + {S}_{B} + 2{S}_{C} & {S}_{B} & 2{S}_{A} + {S}_{B} + {S}_{C} \\ {S}_{A} + 2{S}_{B} + {S}_{C} & 2{S}_{A} + {S}_{B} + {S}_{C} & {S}_{C} \en... | Yes |
Lemma 11. The infinite points of the Kiepert hyperbola are\n\n\[ \left( {\left( {{b}^{2} - {c}^{2}}\right) \left( {{\mathrm{P}}_{a} + \varepsilon \sqrt{\mathrm{Q}}}\right) : \left( {{c}^{2} - {a}^{2}}\right) \left( {{\mathrm{P}}_{b} + \varepsilon \sqrt{\mathrm{Q}}}\right) : \left( {{a}^{2} - {b}^{2}}\right) \left( {{\m... | The point for \( \varepsilon = - 1 \) is the infinite point of the minor axis of the Steiner ellipse. It is the point \( {X}_{3414} \) of [2]. The infinite point corresponding to \( \varepsilon = + 1 \) is \( {X}_{3413} \) . | Yes |
Proposition 12. The two real common points of the conics \( {F}_{a} = 0,{F}_{b} = 0,{F}_{c} = 0 \) are\n\n\[ \n{P}_{\varepsilon } = \left( {{\varepsilon t} + \left( {{b}^{2} - {c}^{2}}\right) \left( {{\mathrm{P}}_{a} - \sqrt{\mathrm{Q}}}\right) : \cdots : \cdots }\right) \n\]\n\n(16)\n\nfor \( \varepsilon = \pm 1 \), w... | Proof. A point on the minor axis of the Steiner ellipse is of the form (16) for some \( t \) . Substituting into the equation \( {F}_{a} = 0 \), or equivalently (15), and solving, we obtain\n\n\[ \n{t}^{2} = \frac{1}{9}\left( {\lambda + \mu \sqrt{Q}}\right) \n\]\n\nwhere\n\n\[ \n\lambda = 4{\mathrm{Q}}^{2} + {\mathrm{{... | Yes |
Proposition 3. Under each lower arc of an \( f \) -belos, construct a new \( f \) -belos similar to the original. Of the four lower arcs, the middle two are congruent, and their common length equals one half the harmonic mean of the lengths of the original lower arcs. | Proof. It is enough to apply the previous lemma noting that the lengths of the considered arcs are proportional to the length of the horizontal segment that they determine. | No |
Proposition 4. Given an \( f \) -belos, the parallelogram \( \mathcal{P}\left( {x}_{0}\right) \) is a rectangle if and only if \( f{\left( {x}_{0}\right) }^{2} = {x}_{0} - {x}_{0}^{2} \) . Consequently, \( \mathcal{P}\left( {x}_{0}\right) \) is a rectangle for every \( {x}_{0} \in \left( {0,1}\right) \) if and only if ... | Proof. We have that \( \overrightarrow{{P}_{2}P} = \left( {p - p{x}_{0}, - {pf}\left( {x}_{0}\right) }\right) \) . Then, \( \mathcal{P}\left( {x}_{0}\right) \) is a rectangle if and only if \( \overrightarrow{{P}_{2}{P}_{1}} \bot \overrightarrow{{P}_{2}P} \) ; i.e., if and only if\n\n\[ 0 = \overrightarrow{{P}_{2}P} \c... | Yes |
Proposition 5. Given an \( f \) -belos, let \( c \in \left( {0,1}\right) \) be such that \( f\left( c\right) \) is the mean value of \( f \) on \( \left\lbrack {0,1}\right\rbrack \) and let \( \mathcal{P}\left( c\right) \) be the parallelogram associated to \( c \) (see Figure 4) . Then: | Proof. Let us denote by \( A \) the area below the upper arc of the \( f \) -belos; i.e.,\n\n\[ A = {\int }_{0}^{1}f\left( x\right) {dx} \]\n\nBy similarity, the area below the lower arc corresponding to the graph of \( g \) is \( {p}^{2}A \) , while the area below the lower arc corresponding to the arc of \( h \) is \... | Yes |
Proposition 6. The area of \( \mathcal{T} \) is given by:\n\n\[ \operatorname{area}\left( \mathcal{T}\right) = p\left( {1 - p}\right) \left| \frac{{f}^{\prime }\left( 0\right) {f}^{\prime }\left( 1\right) }{{f}^{\prime }\left( 0\right) - {f}^{\prime }\left( 1\right) }\right| . \] | Proof. Since the sides of the parallelogram are given by the equations\n\n\[ \overline{{T}_{1}{T}_{2}} : y = {f}^{\prime }\left( 0\right) x \]\n\n\[ \overline{P{T}_{3}} : y = {f}^{\prime }\left( 0\right) \left( {x - p}\right) ,\]\n\n\[ \overline{{T}_{1}P} : y = {f}^{\prime }\left( 1\right) \left( {x - p}\right) ,\]\n\n... | Yes |
Proposition 7. The tangent parallelogram \( \mathcal{T} \) is a rectangle if and only if \( {f}^{\prime }\left( 0\right) {f}^{\prime }\left( 1\right) = \) -1 . In such case we have that\n\n\[ \operatorname{area}\left( \mathcal{T}\right) = p\left( {1 - p}\right) \frac{{f}^{\prime }\left( 0\right) }{1 + {f}^{\prime }{\le... | Proof. The first part is straightforward recalling that two lines of slopes \( {m}_{1} \) and \( {m}_{2} \) are perpendicular if and only if \( {m}_{1}{m}_{2} = - 1 \) . For the second statement it is enough to put \( {f}^{\prime }\left( 1\right) = - 1/{f}^{\prime }\left( 0\right) \) in the previous property. | No |
Proposition 8. Given an \( f \) -belos, let \( f\left( c\right) \) be the mean value of \( f \) on \( \left\lbrack {0,1}\right\rbrack \) . Let also \( \mathcal{P}\left( c\right) \) be the parallelogram associated to \( c \) and \( \mathcal{T} \) the tangent parallelogram. If \( {f}^{\prime }\left( 0\right) {f}^{\prime ... | Proof. Properties 4 and 6 imply that\n\n\[ \operatorname{area}\left( {\mathcal{P}\left( c\right) }\right) = f\left( c\right) p\left( {1 - p}\right) = f\left( c\right) \frac{\operatorname{area}\left( \mathcal{T}\right) }{\frac{{f}^{\prime }\left( 0\right) }{1 + {f}^{\prime }{\left( 0\right) }^{2}}} \geq {2f}\left( c\rig... | Yes |
Proposition 9. Given an \( f \) -belos such that its tangent parallelogram \( \mathcal{T} \) is a rectangle; i.e., such that \( {f}^{\prime }\left( 0\right) {f}^{\prime }\left( 1\right) = - 1 \), the circumcircle \( \Gamma \) of \( \mathcal{T} \) intersects the axis\n\n\( {OX} \) at the point \( \left( {p,0}\right) \) ... | Proof. The center of this circle is the midpoint of \( \overline{{T}_{1}{T}_{3}} \) ; i.e.,\n\n\[ \n{C}_{\Gamma } = \left( {\frac{p + 1 + p{f}^{\prime }{\left( 0\right) }^{2}}{2\left( {1 + {f}^{\prime }{\left( 0\right) }^{2}}\right) },\frac{{f}^{\prime }\left( 0\right) }{2\left( {1 + {f}^{\prime }{\left( 0\right) }^{2}... | Yes |
The point \( \left( {p, f\left( p\right) }\right) \) lies on \( \overline{{T}_{1}{T}_{3}} \) (see Figure 7) if and only if\n\n\[ f\left( p\right) = \frac{p\left( {p - 1}\right) {f}^{\prime }\left( 0\right) {f}^{\prime }\left( 1\right) }{p{f}^{\prime }\left( 0\right) - \left( {1 - p}\right) {f}^{\prime }\left( 1\right) ... | Proof. For the first part it is enough to substitute \( \left( {p, f\left( p\right) }\right) \) in the equation of \( \overline{{T}_{1}{T}_{3}} \) . | No |
Proposition 11. Given an \( f \) -belos, let \( \mathcal{T} \) be its tangent parallelogram. Then the diagonal of \( \mathcal{T} \) opposite to the cusp is tangent to \( f \) at \( \left( {p, f\left( p\right) }\right) \) for every \( p \in \left( {0,1}\right) \) if and only if \( f \) is a parabola \( f\left( x\right) ... | Proof. The first part of the previous property leads to\n\n\[ f\left( x\right) = \frac{x\left( {x - 1}\right) {f}^{\prime }\left( 0\right) {f}^{\prime }\left( 1\right) }{x{f}^{\prime }\left( 0\right) - \left( {1 - x}\right) {f}^{\prime }\left( 1\right) } \]\n\nfor every \( x \in \left\lbrack {0,1}\right\rbrack \) . In ... | Yes |
For each point \( X \) of the circle \( c \) the triangle \( {XYZ} \) is isosceles and the circle \( {k}_{1}\left( {Y,\left| {XY}\right| }\right) \) is tangent to the circle \( c \) at \( X \) and to line \( h \) at \( Z \). | In fact, the first claim follows from an easy angle chasing argument (See Figure 5). From this, or considering the locus as the image \( f\left( c\right) \) of the circle via the homology, follows also easily that the geometric locus of the centers \( Y \) of the circles \( {k}_{1} \) is a parabola tangent to the trian... | No |
Lemma 3. Circle \( {c}^{\prime } \) is orthogonal to circles \( {k}_{1} \) and \( {k}_{2} \) and points \( X, Z,{X}^{\prime }{Z}^{\prime } \) are concyclic. | This follows by considering the inversion with respect to circle \( {c}^{\prime } \) . It is easily seen that this interchanges circle \( c \) and line \( h \) and fixes, as a whole, circles \( {k}_{1} \) and \( {k}_{2} \) . Hence these two circles are orthogonal to \( {c}^{\prime } \) . The second claim is a consequen... | Yes |
Lemma 4. The lines \( Y{Y}^{\prime } \), joining the centers of the circles \( {k}_{1},{k}_{2} \), pass through a fixed point I lying on line \( {OP} \) . The points \( P, I \) are related by the homology \( f \) defined by the triple \( \left( {c, h, P}\right) \) : | \[ \frac{PO}{PS} = \frac{IO}{IS} \cdot \frac{DO}{DE} \] In fact, lines \( Y{Y}^{\prime } \) are images \( Y{Y}^{\prime } = f\left( {X{X}^{\prime }}\right) \) of lines \( X{X}^{\prime } \) under the homology \( f \), hence they pass all through \( I = f\left( P\right) \) . As a result, \( I \) lies on \( {OP} \) and tak... | Yes |
Theorem 5. With the previous notations the map \( {Z}^{\prime } = f\left( Z\right) \) on line \( h \) is an involution, which in line coordinates with origin at the projection \( Q \) of \( I \) on \( h \) obtains the form, for a constant \( w \) : | \[ {z}^{\prime } = \frac{-{w}^{2}}{z} \] | Yes |
Theorem 7. The following properties of the basic configuration are true.\n\n(1) The radical axis \( {DN} \) of circles \( {k}_{1} \) and \( {k}_{2} \) passes through \( D \) and the middle \( N \) of \( Z{Z}^{\prime } \) . | Property (1) immediately follows from Lemma 3, since then \( D \) is on the radical axis of the two circles, and because of the tangency to line \( h \), the middle \( N \) of \( Z{Z}^{\prime } \) is also on the radical axis of \( {k}_{1} \) and \( {k}_{2} \) (See Figure 10). | Yes |
Corollary 8. The radical axis of circle \( {m}_{X} \) and \( c \) passes through a fixed point \( R \) on \( {V}_{1}{V}_{2} \) and passes also through \( {H}_{X} \) . | The first claim is a well known property for the radical axis of members of a pencil and a fixed circle ([2, p.210]). The second claim follows from the fact that \( {H}_{X} \) is the radical center of circles \( {c}^{\prime }, c \) and \( {m}_{X} \) . | Yes |
Theorem 9. The following properties are valid.\n\n(1) The circle \( {n}_{X} \) with diameter \( Z{Z}^{\prime } \) is orthogonal to \( {k}_{1} \) and \( {k}_{2} \) . | Property (1) follows from the fact that \( {n}_{X} \) has its center at the middle \( N \) of \( Z{Z}^{\prime } \) , from which the tangents to \( {k}_{1},{k}_{2} \) are equal (See Figure 11). | Yes |
Corollary 10. The radical axis of circle \( {n}_{X} \) and \( c \) passes through a fixed point \( T \) on \( W{W}^{\prime } \), and passes also through point \( {H}_{X} \) . | The first claim is a well known property for the radical axis of members of a pencil and a fixed circle ([2, p.210]). The second claim follows from the fact that \( {H}_{X} \) is the radical center of circles \( c,{m}_{X} \) and \( {n}_{X} \) . | Yes |
Theorem 11. Lines \( Z{W}_{1} \) and \( {Z}^{\prime }{W}_{2} \) are parallel and \( {W}_{X} \) varies on a fixed line \( {h}^{\prime \prime } \) parallel to \( h \) at distance equal to \( \left| {WI}\right| \) . | To prove the theorem a short calculation seems unavoidable. For this, we use coordinates along line \( h \) and its orthogonal through \( Q \), as in \( §3 \) . Denoting coordinates with respective small letters and by \( r,{r}^{\prime },{r}_{1},{r}_{2} \) respectively the radii of circles \( c,{c}^{\prime },{k}_{1},{k... | Yes |
Theorem 13. The tangents to circles \( {k}_{1},{k}_{2} \), which are reflections of \( h \), respectively, on \( {W}_{X}{W}_{1},{W}_{X}{W}_{2} \) are parallel and equidistant to the median \( {W}_{X}M \) of the right angled triangle \( {W}_{X}{W}_{1}{W}_{2} \) . This median passes through a fixed point \( A \) , indepe... | That the tangents are parallel follows easily by measuring the angles at their intersection points with \( h \) . Similarly follows their parallelity to the median \( {W}_{X}M \) (See Figure 14).\n\nTo show that line \( {W}_{X}M \) passes through a fixed point it suffices, according to the previous lemma, to show that ... | Yes |
Lemma 15. For \( I \) varying on circle \( {c}^{\prime } \) the corresponding point \( P \) varies on an ellipse, which is tangent to circle \( c \) at points \( B \) and \( C \) . | In fact, since \( I = f\left( P\right) \) is a homology, the claimed relation is given through the inverse homology \( P = {f}^{-1}\left( I\right) \), which has the same center and axis and ratio \( {k}^{\prime } = \frac{DE}{DO} \) . Hence the curve \( {c}^{\prime \prime } = {f}^{-1}\left( {c}^{\prime }\right) \) is a ... | No |
Lemma 16. The circle \( {c}^{\prime \prime } \) with diameter \( {Z}_{1}{Z}_{2} \) is orthogonal to circles \( {k}_{1},{k}_{2} \) and the diametral point \( {I}^{\prime } \) defines line \( I{I}^{\prime } \) which is parallel to \( {BC} \) . Also the triples of points \( \left( {{Z}_{1},{I}^{\prime }, V}\right) \) and ... | The proof of parallelity follows from the equality of angles \( {W}_{X}{Z}^{\prime }{Z}_{2},{W}_{X}{Z}_{2}{Z}^{\prime } \) , \( {TI}{Z}_{2} \) (See Figure 17). The collinearity of \( {Z}_{1},{I}^{\prime }, V \) follows from the equality of angles \( {W}_{X}{Z}^{\prime }I, Z{W}_{X}Y \) and the parallelity of \( V{Z}_{1}... | No |
Lemma 18. Line \( {I}^{\prime }{I}^{\prime \prime } \) is orthogonal to \( I{I}^{\prime \prime } \) and passes through point \( {W}_{X} \) and also through points \( G,{G}^{\prime } \), which are correspondingly diametrals of \( I \) with respect to the circle with diameter \( Z{Z}^{\prime } \) and the circle \( {c}^{\... | In fact, by the previous lemma, angle \( {I}^{\prime }{I}^{\prime \prime }I \) is a right one (See Figure 18). Also \( {I}^{\prime \prime } \) is on the circle with diameter \( Y{W}_{X} \), which passes through \( Z,{Z}_{1} \). Thus \( {I}^{\prime },{I}^{\prime \prime },{W}_{X} \) are collinear. The center \( N \) of t... | Yes |
Corollary 19. The cross ratio \( \kappa = \left( {{BC}{W}_{X}{H}_{X}}\right) \) is constant and equal to \( \frac{UB}{UC} \) . Here \( U \) is the projection of \( {G}^{\prime } \) on \( {BC} \) . | This follows by considering the pencil \( {I}^{\prime \prime }\left( {B, C, W,{H}_{X}}\right) \) of lines through \( {I}^{\prime \prime } \) . The lines of the pencil pass through fixed points of the circle \( {c}^{\prime } \) . Thus their cross ratio is independent of the position of \( {I}^{\prime \prime } \) on this... | Yes |
Corollary 20. Fixing point \( I \), which becomes the incenter of triangle \( {ABC} \), the pivoting lines \( A{W}_{X} \) of the Sawayama pairing take the positions of all cevians through \( A \) and Sawayama’s theorem is true. | In fact, by the previous corollary, the constancy of cross ratio implies that for variable \( X \) on \( c \) points \( {H}_{X} \) obtain all positions on line \( {BC} \) and consequently \( {W}_{X} \) , being related to \( {H}_{X} \) by a line-homography, obtains also all possible positions on this line. | No |
Theorem 1. Let \( {U}_{b} = \varphi \left( {{AD},{BC}}\right) \) . The following statements are equivalent.\n\n(a) \( m = \sqrt{2} \), and\n\n(b) \( {U}_{b} = 1 \) . | Proof. We shall use analytic geometry, which offers a simple proof. Let the origin of the rectangular coordinate system be the midpoint \( O \) of the side \( {AB} \) so that the points \( A \) and \( B \) have coordinates \( \left( {-r,0}\right) \) and \( \left( {r,0}\right) \) for some positive real number \( r \) (t... | Yes |
Theorem 2. Let \( {U}_{c} = \varphi \left( {{A}^{\prime }D,{B}^{\prime }C}\right) \) and \( {V}_{c} = \varphi \left( {{A}^{\prime }{D}^{\prime },{B}^{\prime }{C}^{\prime }}\right) \) . Consider the statements (c) \( {U}_{c} = 2,\;\left( {c}^{\prime }\right) \;{V}_{c} = 2,\;\left( {c}^{\prime \prime }\right) \;{U}_{c} =... | Proof. With straightforward computations one can easily check that \[ {V}_{c} - {U}_{c} = \frac{2\left( {{m}^{2} - 2}\right) v{t}^{3}}{{\vartheta }^{2}{\eta }^{2}} \] \[ 2 - {U}_{c} = \frac{\left( {{m}^{2} - 2}\right) v\left( {{2z} + v}\right) }{{m}^{2}{\vartheta }^{2}} \] \[ {V}_{c} - 2 = \frac{\left( {{m}^{2} - 2}\ri... | Yes |
Theorem 3. Let \( {U}_{d} = \varphi \left( {{NG},{NH}}\right) ,{V}_{d} = \varphi \left( {N{G}^{\prime }, N{H}^{\prime }}\right) \) , \( {U}_{e} = \varphi \left( {{OG},{OH}}\right) \) and \( {V}_{e} = \varphi \left( {O{G}^{\prime }, O{H}^{\prime }}\right) \) (see Figure 4).\n\nThe following statements are equivalent.\n\... | Proof. This time the differences \( {U}_{d} - \frac{3}{4} \) and \( {U}_{e} - \frac{1}{4} \) both simplify to \( \frac{{t}^{2}\left( {{m}^{2} - 2}\right) }{4{\vartheta }^{2}} \) , which has the factor \( {m}^{2} - 2 \) again. Similarly, \( {V}_{d} - {U}_{d} \) and \( {V}_{e} - {U}_{e} \) both are equal \( \frac{{mv}{t}... | Yes |
Theorem 4. Let \( {m}_{s} = \frac{{s}^{2} + 2}{4{\left( s + 1\right) }^{2}},{n}_{s} = \frac{3{s}^{2}}{4{\left( s + 1\right) }^{2}} \) ,\n\n\( {U}_{f} = \varphi \left( {O{G}_{s}, O{H}_{s}}\right) ,{V}_{f} = \varphi \left( {O{G}_{s}^{\prime }, O{H}_{s}^{\prime }}\right) ,\n\n\( {U}_{g} = \varphi \left( {N{G}_{s}, N{H}_{s... | Proof. Since \( {G}_{s} = \left( {\frac{-{rst}\left( {m + t}\right) }{\left( {s + 1}\right) \vartheta },\frac{r}{s + 1}}\right) \) and \( {H}_{s} = \left( {\frac{{rs}\left( {1 + z}\right) }{\left( {s + 1}\right) \vartheta },\frac{r}{s + 1}}\right) \), the difference \( {U}_{f} - {m}_{s} \) is \( \frac{{s}^{2}{t}^{2}\le... | Yes |
Theorem 5. Let \( \lambda = \frac{3{s}^{2}}{{s}^{2} + 2} \) . If \( s \neq 0,1 \), then the following statements are equivalent.\n\n(a) \( m = \sqrt{2} \) ,\n\n(f*) \( \;{U}_{g} = \lambda {U}_{f} \) ,\n\n( \( {\mathrm{g}}^{ * } \) ) \( \;{V}_{g} = \lambda {V}_{f} \) . | Proof. For \( s \neq 0,1 \), the equivalence of \( \left( a\right) \) and \( \left( {f}^{ * }\right) \) is a consequence of the equality \( \lambda {U}_{f} - {U}_{g} = \frac{{s}^{2}{t}^{2}\left( {s - 1}\right) \left( {{m}^{2} - 2}\right) }{2\left( {s + 1}\right) \left( {{s}^{2} + 2}\right) {\vartheta }^{2}} \) . | Yes |
Theorem 6. Let \( {U}_{h} = \varphi \left( {B{N}_{1}, A{N}_{2}}\right) ,{V}_{h} = \varphi \left( {B{N}_{3}, A{N}_{4}}\right) \) , \( {U}_{i} = \varphi \left( {N{N}_{1}, N{N}_{2}}\right) ,{V}_{i} = \varphi \left( {N{N}_{3}, N{N}_{4}}\right) \) (see Figure 4).\n\nThe following statements are equivalent.\n\n(a) \( m = \sq... | Proof. Since \( {N}_{1} = \left( {\frac{-{rt}\left( {m + t}\right) }{\vartheta },\frac{r}{\vartheta }}\right) \) and \( {N}_{2} = \left( {\frac{r\left( {1 + z}\right) }{\vartheta },\frac{r{t}^{2}}{\vartheta }}\right) \), we get\n\n\[ \n{U}_{h} - \frac{3}{2} = {U}_{i} - \frac{1}{2} = \frac{{t}^{2}\left( {{m}^{2} - 2}\ri... | Yes |
Theorem 7. The following statements are equivalent.\n\n(a) \( m = \sqrt{2} \) ,\n\n(j) \( \left| {{N}_{1}{N}_{2}}\right| = \left| {AN}\right| ,\;\left( {\mathrm{j}}^{\prime }\right) \;\left| {{N}_{3}{N}_{4}}\right| = \left| {AN}\right| ,\;\left( {\mathrm{j}}^{\prime \prime }\right) \;\left| {{N}_{1}{N}_{2}}\right| = \l... | Proof. For \( \left( a\right) \Leftrightarrow \left( j\right) \), we easily get \( {\left| {N}_{1}{N}_{2}\right| }^{2} - {\left| AN\right| }^{2} = \frac{2{r}^{2}{t}^{2}\left( {{m}^{2} - 2}\right) }{{\vartheta }^{2}} \) .\n\nSince the coordinates of \( {N}_{3} \) and \( {N}_{4} \) are \( \left( {\frac{{rt}\left( {m - t}... | Yes |
Theorem 8. Let \( {U}_{\ell } = \varphi \left( {A{N}_{5}, B{N}_{6}}\right) ,{V}_{\ell } = \varphi \left( {A{N}_{7}, B{N}_{8}}\right) \) ,\n\n\( {U}_{m} = \varphi \left( {G{N}_{6}, H{N}_{5}}\right) ,{V}_{m} = \varphi \left( {{G}^{\prime }{N}_{8},{H}^{\prime }{N}_{7}}\right) \) and\n\n\( {U}_{n} = \varphi \left( {N{N}_{5... | Proof. Since \( {N}_{5} = \left( {\frac{-r{t}^{2}}{\vartheta },\frac{-r\left( {1 + z}\right) }{\vartheta }}\right) \) and \( {N}_{6} = \left( {\frac{r}{\vartheta },\frac{-{rt}\left( {m + t}\right) }{\vartheta }}\right) \), we obtain \( {U}_{\ell } - \frac{1}{2} = \frac{{t}^{2}\left( {{m}^{2} - 2}\right) }{2{\vartheta }... | Yes |
Theorem 9. Let \( {U}_{o} = \left| {N{M}_{1}}\right| - \left| {AN}\right| ,{V}_{o} = \left| {N{M}_{2}}\right| - \left| {AN}\right| \) ,\n\n\( {U}_{p} = \varphi \left( {{M}_{1}{N}_{1},{M}_{1}{N}_{2}}\right) \) and \( {V}_{p} = \varphi \left( {{M}_{2}{N}_{3},{M}_{2}{N}_{4}}\right) \) (see Figure 5).\n\nThe following stat... | Proof. Since \( {M}_{1} = \left( {\frac{ru}{\vartheta }, - \frac{rz}{\vartheta }}\right) \), the difference \( {\left| {M}_{1}N\right| }^{2} - {\left| AN\right| }^{2} \) is \( \frac{2{r}^{2}{t}^{2}\left( {{m}^{2} - 2}\right) }{{\vartheta }^{2}} \) . Similarly, \( {\left| {M}_{2}N\right| }^{2} - {\left| {M}_{1}N\right| ... | Yes |
Theorem 10. Let \( {U}_{q} = \varphi \left( {{G}_{2}{G}_{1},{G}_{2}{G}_{3}}\right) \) and \( {V}_{q} = \varphi \left( {{G}_{5}{G}_{4},{G}_{5}{G}_{6}}\right) \) . The following statements are equivalent: (a) \( m = \sqrt{2} \), (q) \( {U}_{q} = \frac{1}{9},\;\left( {\mathrm{q}}^{\prime }\right) \;{V}_{q} = \frac{1}{9},\... | Proof. If \( X = \left( {x, y}\right) \), then the points \( {G}_{1},{G}_{2} \) and \( {G}_{3} \) have the same ordinate \( \frac{y}{3} \) while their abscissae are \( \frac{x}{3} - \frac{{2rt}\left( {m + t}\right) }{3\vartheta },\frac{x}{3} + \frac{2ru}{3\vartheta } \) and \( \frac{x}{3} + \frac{{2r}\left( {z + 1}\rig... | Yes |
Theorem 11. Let \( {U}_{s} = \varphi \left( {{NU},{NV}}\right) ,{U}_{t} = \varphi \left( {{OU},{OV}}\right) \) and \( {V}_{t} = \varphi \left( {{N}_{6}U,{N}_{5}V}\right) \) . The following statements are equivalent:\n\n(a) \( m = \sqrt{2} \) ,\n\n(s) \( {U}_{s} = 1 \) ,\n\n(t) \( {U}_{t} = \frac{1}{2},\;\left( {\mathrm... | Proof. Since abscissae of \( U \) and \( V \) are \( \frac{r\left( {{uv} + {z}^{2}}\right) }{\eta \vartheta } \) and \( \frac{r\left( {{uv} - {z}^{2}}\right) }{\eta \vartheta } \), we get \( {U}_{s} - 1 = \) \( {U}_{t} - \frac{1}{2} = \frac{{t}^{2}{v}^{2}\left( {{m}^{2} - 2}\right) }{{\eta }^{2}{\vartheta }^{2}} \) . T... | Yes |
Theorem 12. Let \( W = U \oplus V \) , \( {U}_{u} = \left| {WO}\right| - \frac{\left| AB\right| }{2} \) , \( {U}_{v\left( {i, j}\right) } = \varphi \left( {W{N}_{i}, W{N}_{j}}\right) \) for \( i \in \{ 1,3\} \) and \( j \in \{ 2,4\} \) and \( {U}_{w} = \varphi \left( {{W}^{ * }N,{WN}}\right) \) . The following statemen... | Proof. Since the coordinates of the point \( W \) is the pair \( \left( {\frac{ruv}{\eta \vartheta },\frac{r{z}^{2}}{\eta \vartheta }}\right) \), we get that \( {\left| WO\right| }^{2} - {r}^{2} = \frac{2{r}^{2}{t}^{2}{v}^{2}\left( {{m}^{2} - 2}\right) }{{\eta }^{2}{\vartheta }^{2}} \) . This proves \( \left( a\right) ... | Yes |
Theorem 13. Let \( \lambda = \frac{7}{4} + \sqrt{2},{U}_{z} = \varphi \left( {{A}^{\prime }{K}_{2},{B}^{\prime }{K}_{1}}\right) \) and \( {V}_{z} = \varphi \left( {{A}^{\prime }{K}_{4},{B}^{\prime }{K}_{3}}\right) \) . For the statements (a) \( m = \sqrt{2} \) , (z) \( {U}_{z} = \lambda ,\;\left( {\mathrm{z}}^{\prime }... | Proof. Since \( {K}_{2} \) and \( {K}_{1} \) have abscissae \( - \frac{r{t}^{2}}{\vartheta } \) and \( \frac{r}{\vartheta } \), we conclude that \( \lambda - {U}_{z} = \) \( \frac{\left\lbrack {\left( {4\left( {m + 1}\right) {\vartheta }^{2} - {z}^{2}}\right) \sqrt{2} + m\left( {z + {2v}}\right) \left( {{3z} + {2v}}\ri... | Yes |
Theorem 14. Let \( \lambda = 1 + \frac{\sqrt{2}}{2},{U}_{\alpha } = \varphi \left( {A{K}^{ * },{BL}}\right) \) and \( {V}_{\alpha } = \varphi \left( {{AS}, B{T}^{ * }}\right) \) . The following statements are equivalent: (a) \( m = \sqrt{2} \), ( \( \alpha \) ) \( \;{U}_{\alpha } = \lambda ,\;\left( {\alpha }^{\prime }... | Proof. Since \( \left( {\frac{r\left( {\vartheta - {tz}}\right) }{mv}, - \frac{r\left( {m + \eta }\right) }{mv}}\right) \) and \( \left( {\frac{r\left( {m - \vartheta }\right) }{mv}, - \frac{r\left( {{tz} + \eta }\right) }{mv}}\right) \) are the coordinates of the points \( {K}^{ * } \) and \( L \), we get that \( \lam... | Yes |
Theorem 15. Let \( {\lambda }_{ \pm } = 1 \pm \frac{\sqrt{2}}{2} \), \( {U}_{\beta } = \varphi \left( {N{S}_{1}, N{T}_{1}}\right) ,{V}_{\beta } = \varphi \left( {N{S}_{2}, N{T}_{2}}\right) , {U}_{\gamma } = \varphi \left( {{N}^{ * }{S}_{1},{N}^{ * }{T}_{1}}\right) ,{V}_{\gamma } = \varphi \left( {{N}^{ * }{S}_{2},{N}^{... | Proof. For the implication \( \left( a\right) \Rightarrow \left( \delta \right) \), from the right-angled triangles \( {OG}{S}_{1} \) and \( {OH}{T}_{1} \) and Theorem 3, we get that the sum \( O{S}_{1}^{2} + O{T}_{1}^{2} \) is equal to \( \left( {O{G}^{2} + G{S}_{1}^{2}}\right) + \) \( \left( {O{H}^{2} + H{T}_{1}^{2}}... | Yes |
Theorem 16. Let \( {U}_{\varepsilon } = \varphi \left( {{A}^{\prime }Q,{B}^{\prime }R}\right) \) and \( {V}_{\varepsilon } = \varphi \left( {{A}^{\prime }{Q}^{\prime },{B}^{\prime }{R}^{\prime }}\right) \) . Consider the statements (a) \( m = \sqrt{2} \) , (ε) \( \;{U}_{\varepsilon } = 5,\;\left( {\varepsilon }^{\prime... | Proof. Since \( Q = \left( {\frac{r\left( {u + {3z} + 2}\right) }{\vartheta },0}\right) \) and \( R = \left( {\frac{r\left( {{3u} - {3z} - 2}\right) }{\vartheta },0}\right) \), we get that \( 5 - {U}_{\varepsilon } \) is equal to \( \frac{\eta \left( {{m}^{2} - 2}\right) \left( {\vartheta + {2z}}\right) }{{m}^{2}{\vart... | Yes |
Theorem 17. (i) The triangles \( {ADP} \) and \( {BCP} \) have the same orthocenter that lies on a circle with the segment \( {CD} \) as a diameter. | Proof. The orthocenters of the triangles \( {ADP} \) and \( {BCP} \) both have coordinates \( \left( {\frac{ru}{v},\frac{{2mr}{t}^{2}}{\vartheta v}}\right) \). If \( {K}_{0} \) is the midpoint of the segment \( {CD} \) and \( {L}_{0} \) is the orthocenter of the triangle \( {ADP} \), then \( {\left| C{K}_{0}\right| }^{... | Yes |
Theorem 18. The equalities\n\n\[ \n\varphi \left( {{AK}, B{L}^{ * }}\right) = \varphi \left( {A{S}^{ * },{BT}}\right) \;\text{ and }\;\varphi \left( {A{K}^{ * },{BL}}\right) = \varphi \left( {{AS}, B{T}^{ * }}\right) \n\]\n\nhold for all points \( P \) on the circle and every ratio \( m \) . | Proof. Both \( \varphi \left( {{AK}, B{L}^{ * }}\right) \) and \( \varphi \left( {A{S}^{ * },{BT}}\right) \) have the value \( \frac{{\left( m - 1\right) }^{2} + 1}{2{m}^{2}} \) while both \( \varphi \left( {A{K}^{ * },{BL}}\right) \) and \( \varphi \left( {{AS}, B{T}^{ * }}\right) \) have the value \( \frac{{\left( m ... | Yes |
Theorem 19. The ratio \( m \) is either \( \sqrt{2} \) or \( 2 + \sqrt{2} \) if and only if for every point \( P \) on the circle the equalities \( \varphi \left( {{AK}, B{L}^{ * }}\right) = \lambda \) and/or \( \varphi \left( {A{S}^{ * },{BT}}\right) = \lambda \) hold. | Proof. Since \( \left( {-\frac{r\left( {\vartheta + {tz}}\right) }{mv},\frac{r\left( {m - \eta }\right) }{mv}}\right) \) and \( \left( {\frac{r\left( {m + \vartheta }\right) }{mv},\frac{r\left( {{tz} - \eta }\right) }{mv}}\right) \) are the coordinates of the points \( K \) and \( {L}^{ * } \), we get that \( \lambda -... | Yes |
Theorem 2. [1, p.108] The points \( \mathbf{P} \) for which the pedal triangle \( \mathbf{H} \) is isosceles are all and only the points that lie on at least one of the Apollonius circles associated to the vertices of \( \mathbf{P} \) . | The Apollonius circle associated to the vertex \( {A}_{i} \) is the locus of points \( P \) such that \( P{A}_{i + 1} : P{A}_{i + 2} = {A}_{i}{A}_{i + 2} : {A}_{i}{A}_{i + 1} \) . The three Apollonius circles are coaxial and they intersect in the two isodynamic points of the triangle \( \mathbf{P},{I}_{1} \) and \( {I}... | No |
Theorem 5. If \( \mathbf{P} \) is a quadrilateral which is not a parallelogram, the reflections of the Simson point with respect to the lines containing the sides of \( \mathbf{P} \) are collinear and the line \( \ell \) containing them is parallel to the Simson line. | Proof. The theorem is trivially true if \( \mathbf{P} \) has one pair of parallel sides. Suppose that \( \mathbf{P} \) is without parallel sides (see Figure 7). Let \( {K}_{i}, i = 1,2,3,4 \), be the reflection of \( S \) with respect to the line \( {A}_{i}{A}_{i + 1} \) . The points \( S,{H}_{i} \) and \( {K}_{i} \) a... | Yes |
Theorem 6. The points \( P \) whose pedal quadrilaterals have at least one pair of parallel sides are precisely those on the circles \( {A}_{1}{A}_{3}S \) and \( {A}_{2}{A}_{4}S \) . | In general, the circles \( {A}_{1}{A}_{3}S \) and \( {A}_{2}{A}_{4}S \) intersect at two points, the Simson point \( S \) and one other point \( {P}^{ * } \) (see Figure 10). The pedal quadrilateral of \( {P}^{ * } \) is a parallelogram. We call \( {P}^{ * } \) the parallelogram point of \( \mathbf{P} \) . Observe that... | Yes |
Lemma 7. If \( {ABCD} \) is a convex or a crossed quadrilateral such that \( {ABC} \) and CDA are right angles, then it is cyclic. Moreover, its circumcenter is the midpoint of \( {AC} \) and its anticenter is the midpoint of \( {BD} \) . | Proof. Let \( {ABCD} \) be a convex or a crossed quadrilateral with \( {ABC} \) and \( {CDA} \) right angles (see Figure 11). Then, it is cyclic with the diagonal \( {AC} \) as diameter. (When it is a crossed quadrilateral it is inscribed in the semicircle with diameter \( {AC}) \) . Then its circumcenter \( O \) is th... | Yes |
Theorem 8. The polygon \( {\mathbf{P}}_{c}\left( P\right) \) is the image of \( \mathbf{P} \) under the homothety \( \mathrm{h}\left( {P,\frac{1}{2}}\right) \) . | Proof. By Lemma 7, the circumcenter \( {O}_{i} \) of \( {\mathbf{Q}}_{i} \) is the midpoint of \( {A}_{i}P \) (see Figure 12 for a pentagon \( \mathbf{P} \) ). | No |
Theorem 9. The polygon \( {\mathbf{P}}_{a}\left( P\right) \) is the medial polygon of \( \mathbf{H} \), with vertices the midpoints of the segments \( {H}_{i}{H}_{i + 1} \) for \( i = 1,2,\ldots, n \) . | Proof. By Lemma 7, the anticenter \( {A}_{i}^{\prime } \) of \( {\mathbf{Q}}_{i} \) is the midpoint of \( {H}_{i}{H}_{i + 3} \) (see Figure 13 for a pentagon). | No |
Theorem 11. If \( \mathbf{H} \) is cyclic, the Euler lines of the quadrilaterals \( {\mathbf{Q}}_{i} \) are concurrent at the circumcenter of \( \mathbf{H} \) . | Proof. The Euler line of the quadrilateral \( {\mathbf{Q}}_{i} \) passes through the circumcenter \( {O}_{i} \) of \( {\mathbf{Q}}_{i} \) and through the anticenter \( {A}_{i}^{\prime } \) of \( {\mathbf{Q}}_{i} \), that is the midpoint of \( {H}_{i}{H}_{i + 3} \), then it is the perpendicular bisector of a side of \( ... | No |
Lemma 1. Denote by \( {m}_{E} \) the slope of Euler’s line in \( \bigtriangleup {ABC} \) and by \( {m}_{1},{m}_{2},{m}_{3} \) the slopes of the lines \( {BC},{AC} \), and \( {AB} \), respectively. Then\n\n\[ \n{m}_{E} = - \frac{{m}_{1}{m}_{2} + {m}_{3}{m}_{1} + {m}_{2}{m}_{3} + 3}{{m}_{1} + {m}_{2} + {m}_{3} + 3{m}_{1}... | Proof. Measuring the slope of the angle between \( {BC} \) and the Euler’s line of \( \bigtriangleup {ABC} \) , we have (see Figure 1):\n\n\n\n\[ \n\frac{{m}_{1} - {m}_{E}}{1 + {m}_{1}{m}_{E}} = \tan \angle \left( ... | Yes |
Lemma 2. Euler’s line of \( \bigtriangleup {ABC} \) intersects the lines \( {AB} \) and \( {AC} \) in \( M \), respectively \( N \) . Then Euler’s line of \( \bigtriangleup {AMN} \) is parallel to \( {BC} \) . | Proof. Choose a coordinate system so that the \( x \) -axis is parallel to \( {BC} \), as in Application 3 (see Figure 3). If we denote by \( {m}_{1} \) the slope of the straight line \( {BC} \) , then \( {m}_{1} = 0 \) . Denoting \( {m}_{2},{m}_{3},{m}_{e} \) the slopes of the straight lines \( {AC},{AB} \), and respe... | Yes |
Now we describe two triangles of interest that have the same Euler's line. Consider \( \bigtriangleup {ABC} \) and its circumcircle \( \mathcal{C} \) . Consider also the incircle tangent to \( {BC},{AC} \) and \( {AB} \) respectively in \( D, E \), and \( F \) . On the straight lines \( {AI},{BI},{CI} \) we consider th... | Thus, \( I \) is the orthocenter in \( \Delta {I}_{a}{I}_{b}{I}_{c} \), and \( O \) is the center of the nine-point circle in \( \Delta {I}_{a}{I}_{b}{I}_{c} \) . Therefore, \( {OI} \) is Euler’s line in \( \Delta {I}_{a}{I}_{b}{I}_{C} \) . Remark that \( \bigtriangleup {DEF} \) and \( \Delta {I}_{a}{I}_{b}{I}_{c} \) h... | Yes |
Proposition 3 (Gossard, [15]). Denote by e the Euler line of an arbitrary \( \bigtriangleup {ABC} \) in the Euclidean plane. Suppose that e intersects \( {BC},{AB},{AC} \) in \( M, N \), and respectively \( P \) . Denote by \( {e}_{1},{e}_{2},{e}_{3} \) Euler’s lines of \( {\Delta ANP},{\Delta BMN} \), and \( {\Delta C... | Proof. We choose coordinate axis such that the vertices of \( \bigtriangleup {ABC} \) have the coordinates \( A\left( {0,1}\right), B\left( {b,0}\right), C\left( {c,0}\right) \) (see Figure 4). Let \( G \) be the gravity center of \( {\Delta ABC} \) ; then \( G\left( {\frac{b + c}{3},\frac{1}{3}}\right) \) . The slope ... | Yes |
We have seen in Example 1 (see [19],244) that if \( H \) is the orthocenter of \( \bigtriangleup {ABC} \), and \( {O}_{a},{O}_{b},{O}_{c} \) are the circumcenters of triangles \( {BHC} \) , \( {CHA},{AHB} \), then \( \bigtriangleup {ABC} \) and \( \bigtriangleup {O}_{a}{O}_{b}{O}_{c} \) have the same Euler’s line (see ... | Then \( H \) is the circumcenter of \( \Delta {O}_{a}{O}_{b}{O}_{c} \) . Actually, \( \Delta {O}_{a}{O}_{b}{O}_{c} \) is the homothetic of \( \Delta {A}_{1}{B}_{1}{C}_{1} \) by homothety of center \( O \) and ratio 2 . Thus, \( \Delta {O}_{a}{O}_{b}{O}_{c} \) has the sides parallel and congruent to the sides of \( \big... | Yes |
Proposition 6. Consider \( {A}^{\prime },{B}^{\prime },{C}^{\prime } \) the midpoints of the sides \( {BC},{AC} \) and respectively \( {AB} \) . Let \( {A}_{1},{B}_{1} \) and \( {C}_{1} \) the feet of the altitudes. Consider the projectivity \( \phi \) uniquely determined by \( {A}_{1} \rightarrow {B}_{1},{A}^{\prime }... | The proof is just a direct application of Pappus' Theorem on the circle, for the geometric structure described in the statement. Since \( H \) and \( {\Omega }_{AB} \) are on Euler’s line, the axis of projectivity and Euler's line must be the same straight line. As a consequence, the third point, \( {O}_{9} \), the cen... | Yes |
Theorem 1. If \( {a}_{\mathrm{o}} \) and \( {a}_{\mathrm{i}} \) are the lengths of the sides of \( {\mathbf{P}}_{\mathrm{o}} \) and \( {\mathbf{P}}_{\mathrm{i}} \) respectively, then\n\n\[ {a}_{\mathrm{o}} = {a}_{\mathrm{i}}\left( {1 + \sqrt{1 + {\left( \sin \frac{\pi }{n}\right) }^{-2}}}\right) \]\n\n(1) | ## 3. Proof of Theorem 1\n\nIf \( {r}_{\mathrm{i}} \) is the radius of the incircle of \( {\mathbf{P}}_{\mathrm{i}} \), then \( {r}_{\mathrm{i}} = \frac{1}{2}{a}_{\mathrm{i}}\cot \theta \) . Let \( {F}_{\mathrm{o}},{F}_{\mathrm{i}} \) and \( F \) be the areas of \( {\mathbf{P}}_{\mathrm{o}},{\mathbf{P}}_{\mathrm{i}} \)... | Yes |
Theorem 2. The angle \( \psi \) between \( {a}_{\mathrm{o}} \) and \( v \) is given by\n\n\[ \cos \psi = {\sin }^{2}\frac{\pi }{n} \] | Proof. It follows from (2) that the area of the triangle \( \mathbf{T} \) is given by\n\n\[ F = {\rho }^{2}\cot \frac{\psi }{2} + \rho \left( {v + {a}_{\mathrm{i}}}\right) \]\n\nand\n\n\[ F = {\rho }^{2}\cot \frac{\pi - \left( {{2\theta } + \psi }\right) }{2} + {\rho v}. \]\n\nSubstituting \( {a}_{\mathrm{i}} = {2\rho ... | Yes |
Proposition 1. The area of \( {ABC} \) is the geometric mean of the areas of the triangles \( A{B}_{a}{C}_{a}, B{C}_{b}{A}_{b}, C{A}_{c}{B}_{c} \) . | Proof. From the coordinates of these points, we have\n\n\[ \n\frac{\Delta \left( {A{B}_{a}{C}_{a}}\right) }{\Delta \left( {ABC}\right) } = \frac{{u}^{2}}{vw},\;\frac{\Delta \left( {B{C}_{b}{A}_{b}}\right) }{\Delta \left( {ABC}\right) } = \frac{{u}^{2}}{vw},\;\frac{\Delta \left( {C{A}_{c}{B}_{c}}\right) }{\Delta \left( ... | Yes |
Theorem 2. For any point \( P \), the six points \( {A}_{b},{A}_{c},{B}_{c},{B}_{a},{C}_{a},{C}_{b} \) always lie on a conic. | Proof. They all lie on the conic \( \Gamma \left( P\right) \) with equation\n\n\[ u\left( {v + w}\right) {x}^{2} + v\left( {w + u}\right) {y}^{2} + w\left( {u + v}\right) {z}^{2} + \left( {{vw} + \left( {u + v}\right) \left( {u + w}\right) }\right) {yz} \]\n\n\[ + \left( {{wu} + \left( {v + w}\right) \left( {v + u}\rig... | Yes |
Proposition 3. (b) If the conic \( \Gamma \left( P\right) \) is nondegenerate, it is homothetic to the circumconic with perspector \( \left( {{u}^{2} : {v}^{2} : {w}^{2}}\right) \) and has center \( P \) . | Proof. (b) follows from rewriting the equation of the conic \( \Gamma \left( P\right) \) in the form\n\n\[ \left( {{u}^{2}{yz} + {v}^{2}{zx} + {w}^{2}{xy}}\right) + \left( {x + y + z}\right) \left( {u\left( {v + w}\right) x + v\left( {w + u}\right) y + w\left( {u + v}\right) z}\right) = 0. \] | Yes |
Proposition 4. The conic \( \Gamma \left( P\right) \) and the circumconic with perspector \( {P}^{2} \) have the same infinite points. | \[ \text{If}P = \left( {u : v : w}\right) \text{and}u = v + w\text{, then}\Gamma \left( P\right) \text{factors as} \] \[ \left( {{vx} + {wx} + {vy} + {2wy} + {wz}}\right) \left( {{vx} + {wx} + {vy} + {2vz} + {wz}}\right) = 0, \] two parallel lines to the cevian \( {AP} \) . If \( P \) lies on \( {K327} \), then \( \Gam... | No |
Proposition 6. (a) If \( {A}^{\prime },{B}^{\prime },{C}^{\prime } \) are the midpoints of \( {B}_{a}{C}_{a},{C}_{b}{A}_{b},{A}_{c}{B}_{c} \) respectively, the lines \( A{A}^{\prime }, B{B}^{\prime }, C{C}^{\prime } \) concur at \( P \) . | Proof. In homogeneous barycentric coordinates, \[ {A}^{\prime } = \left( {{2vw} + {wu} + {uv} : - {uv} : - {uw}}\right) , \] These points clearly lie on the line \( {AP} : {wy} - {vz} = 0 \) . Similarly, \( {B}^{\prime },{B}^{\prime \prime } \) lie on \( {BP} \) and \( {C}^{\prime },{C}^{\prime \prime } \) lie on \( {C... | Yes |
Proposition 7. If \( P = \left( {u : v : w}\right) \), both of the triangles formed by the lines \( {B}_{a}{C}_{a} \) , \( {C}_{b}{A}_{b},{A}_{c}{B}_{c} \) and the lines \( {B}_{c}{C}_{b},{C}_{a}{A}_{c},{A}_{b}{B}_{a} \) are perspective with \( {ABC} \) at the isotomic conjugate of the anticomplement of \( P \), i.e., ... | Proof. (a) The lines \( {B}_{a}{C}_{a},{C}_{b}{A}_{b},{A}_{c}{B}_{c} \) have equations \[ {ux} + \left( {w + u}\right) y + \left( {u + v}\right) z = 0, \] \[ \left( {v + w}\right) x + {vy} + \left( {u + v}\right) z = 0, \] \[ \left( {v + w}\right) x + \left( {w + u}\right) y + {wz} = 0. \] They bound a triangle with ve... | Yes |
Lemma 2. Let \( S \) be a point in the interior of two rays \( {AB} \) and \( {AC} \) . Suppose that \( {ABSC} \) is cyclic, and let \( X \) be a point on ray \( {AB} \) such that \( \left| {AX}\right| < \left| {AB}\right| \) . Let \( Y = \left( {AXS}\right) \cap {AC} \) . Then \( \left| {AY}\right| > \left| {AC}\right... | Proof. Since \( \left| {AX}\right| < \left| {AB}\right| ,\angle {AXS} > \angle {ABS} \) . Since \( {ABSC} \) and \( {AXSY} \) are cyclic, \( \angle {ACS} = \pi - \angle {ABS} \) and \( \angle {AYS} = \pi - \angle {AXS} \) . Therefore \( \angle {AYS} < \) \( \angle {ABS} \) so that \( \left| {AY}\right| > \left| {AC}\ri... | Yes |
Lemma 3. Let \( {ABCDEF} \) be a complete quadrilateral where points in each of the triples \( A, B, C;B, D, E \), etc. as in fig. 1 are collinear and angle \( \angle {CDE} \) is obtuse. Denote the Miquel point of ABCDEF by \( S \) . There exist no two points \( X \) and \( Y \) on rays \( {AF},{AB} \) respectively wit... | Proof. The Miquel point \( S \) lies on \( \left( {AFC}\right) \) and \( \left( {ABE}\right) \) . By Lemma 2, no such \( X \) and \( Y \) exist. | No |
Lemma 4. If \( \Pi = {V}_{1}\cdots {V}_{n}, n \geq 5 \) is a convex Simson polygon, then \( \Pi \) has no pair of parallel sides. | Proof. By the nondegeneracy assumption, it is clear that no two consecutive sides can be parallel. So suppose that \( {V}_{1}{V}_{2}\parallel {V}_{i}{V}_{i + 1}, i \notin \{ 1,2, n\} \) . Then \( S \) lies on the Simson line \( L \) orthogonal to \( {V}_{1}{V}_{2} \) and \( {V}_{i}{V}_{i + 1} \) . The projection of \( ... | Yes |
Theorem 5. A convex pentagon does not admit a Simson point. | Proof. Let \( \Pi = {ABCDE} \) be a nondegenerate convex pentagon. Suppose that \( S \) is a point for which the pedal in \( \Pi \) is a line. Then in particular the pedal is a line for every 4 sides of the pentagon. Therefore if \( {BC} \cap {DE} = F \), then \( S \) must be a Simson point for \( {ABFE} \), so that \(... | Yes |
Theorem 6. A convex \( n \) -gon with \( n \geq 5 \) does not admit a Simson point. | Proof. The base case has been established. Assume the hypothesis for \( n \geq 5 \) , and consider the case for an \( \left( {n + 1}\right) \) -gon \( \Pi \) with vertices \( {V}_{1},\ldots ,{V}_{n + 1} \) . Suppose that \( \Pi \) admits a Simson point. Let \( {V}_{n - 1}{V}_{n} \cap {V}_{n + 1}{V}_{1} = {V}^{\prime } ... | Yes |
Theorem 7. An \( n \) -gon \( \Pi = {V}_{1}\cdots {V}_{n} \) admits a Simson point \( S \) if and only if all circles \( \left( {{V}_{i}{W}_{i}{V}_{i + 1}}\right) \) have a common intersection. | Proof. Assume first that \( S \) is a Simson point for \( \Pi \) . The projections of \( S \) into \( {V}_{i - 1}{V}_{i},{V}_{i}{V}_{i + 1} \) and \( {V}_{i + 1}{V}_{i + 2} \) are collinear. By the Simson-Wallace Theorem (Theorem 1), \( S \) is on the circumcircle of \( {V}_{i}{W}_{i}{V}_{i + 1} \) .\n\nConversely, let... | Yes |
Theorem 8. Let \( S \) be a point and \( L \) a line not passing through \( S \) . Suppose that \( {X}_{1},\ldots ,{X}_{n} \) are points on \( L \) such that \( \left| {{X}_{i}{X}_{i + 1}}\right| = \Delta \) for all \( i = 1,\ldots, n - 1 \) and let \( \Pi = {V}_{1}\cdots {V}_{n} \) be the equidistant Simson polygon wi... | Proof. Without loss of generality, let \( S = \left( {0, s}\right), L \) be the \( x \) -axis, \( {X}_{i} = (X + (i - \) \( 1)\Delta ,0) \) and \( {X}_{i + 1} = \left( {X + {i\Delta },0}\right) \) . A calculation shows that the perpendiculars at \( {X}_{i} \) and \( {X}_{i + 1} \) to the segments \( S{X}_{i} \) and \( ... | Yes |
Corollary 9. Let \( C \) be a parabola with focus \( F \) . The locus of projections of \( F \) into the lines tangent to \( C \) is the tangent to \( C \) at its vertex. | Proof. As seen in the proof of Theorem 8, the coordinates of the \( {V}_{i}, i = 1,\ldots, n \) are continuous functions of \( \Delta \) . Therefore as \( n \rightarrow \infty \) and \( \Delta \rightarrow 0 \) in Theorem 8, the limit of the polygon is a parabola with focus \( S \) and tangent line at the vertex equal t... | No |
Theorem 10. The optimal piecewise-continuous linear approximation to \( f\left( x\right) \) with the setup above is given by the sides \( {V}_{1}{V}_{2},{V}_{2}{V}_{3},\ldots ,{V}_{n}{V}_{n + 1} \) of an equidistant Simson \( \left( {n + 2}\right) \) -gon with \( {X}_{1} = \frac{a}{2},\Delta = \frac{b - a}{n} \) and \(... | Proof. The equation of the \( i \) th line segment simplifies to\n\n\[ l\left( x\right) = \frac{x\left( {{x}_{i + 1} + {x}_{i}}\right) - {x}_{i}{x}_{i + 1} - {\Delta }^{2}}{4s},\text{ for }x \in \left\lbrack {{x}_{i},{x}_{i + 1}}\right\rbrack .\n\]\n\nTherefore \( f\left( x\right) - l\left( x\right) = \frac{\left( {x -... | Yes |
Theorem 12. An equidistant Simson polygon \( {V}_{1}{V}_{2}\ldots {V}_{n} \) with projections \( {X}_{1},{X}_{2},\ldots ,{X}_{n} \) has the following properties:\n\n(1) If \( j - i > 0 \) is odd, the segments \( {V}_{i}{V}_{j},{V}_{i + 1}{V}_{j - 1},\ldots ,{V}_{\frac{j + i + 1}{2}}{V}_{\frac{j + i - 1}{2}} \) are para... | Proof. (1). The slope between \( {V}_{i} \) and \( {V}_{j} \) is easily calculated to be \( \frac{{2X} + \left( {i + j - 1}\right) \Delta }{2s} \) .\n\n(2). Recall that the parabola is given by \( y = \frac{{x}^{2} - {\Delta }^{2}}{4s} \) so that its slope at \( {V}_{\frac{j + i}{2}} \) is \( \frac{{2X} + \left( {2\lef... | Yes |
Property 1. Let \( S \) and \( L \) be the Simson point and Simon line of a Simson polygon (not necessarily equidistant) with vertices \( {V}_{1},\ldots ,{V}_{n} \) and define \( {X}_{1},\ldots ,{X}_{n} \) as before. Let \( {V}_{i}^{\prime } \) be the reflection of \( {V}_{i} \) in \( L \) . Then the lines \( {V}_{i}{X... | Proof. The proof is by a straightforward angle count. | No |
Corollary 13. Let \( C \) be a parabola with focus \( F \) and tangent line \( L \) at its vertex. Let \( X \) be any point on \( C \) and \( K \) the tangent at \( X \) . Furthermore, let \( {X}^{\prime } \) be the reflection of \( X \) in \( L \) . Then \( K \) forms equal angles with \( {X}^{\prime }X \) and \( {FX}... | \n\nFigure 6. Corollary 13: \( X \) is a variable point of the parabola \( C, F \) is the focus, \( L \) is the tangent to \( C \) at its vertex and \( {X}^{\prime } \) is the reflection of \( X \) in \( L \) . The... | Yes |
Theorem 14. Let \( {M}_{i} \) be the midpoint of \( {V}_{i}{V}_{i + 1}, i = 1,\ldots, n - 2 \) . Then the midpoints \( {M}_{i} \) lie on a parabola \( {C}^{\prime } \) with focus \( S \) and tangent line at its vertex \( L \) . | Proof. Since \( {V}_{i} = \left( {{2X} + \left( {{2i} - 1}\right) \Delta ,\frac{\left( {x + \left( {i - 1}\right) \Delta }\right) \left( {x + {i\Delta }}\right) }{s}}\right) \) ,\n\n\[ \n{M}_{i} = \left( {2\left( {X + {i\Delta }}\right) ,\frac{{\left( X + i\Delta \right) }^{2}}{s}}\right) .\n\]\n\nTherefore the \( {M}_... | Yes |
Theorem 16. Let \( {W}_{i, j} = {V}_{i}{V}_{i + 1} \cap {V}_{j}{V}_{j + 1} \) for each \( i, j \in \{ 1,2,\ldots, n - 2\} \) and \( i \neq j \) . Let \( {M}_{i, j + 1} \) and \( {M}_{i + 1, j} \) be the respective midpoints of chords \( {V}_{i}{V}_{j + 1} \) and \( {V}_{i + 1}{V}_{j} \) . Then \( {W}_{i, j}{M}_{i, j + ... | Proof. As shown in the proof of Theorem 12, the \( x \) -coordinate of \( {M}_{i, j + 1} \) is \( {2X} + \) \( \left( {i + j}\right) \Delta \) and that of \( {M}_{i + 1, j} \) is the same. The point \( {W}_{i, j} \) is the intersection of the line \( {V}_{i}{V}_{i + 1} \) given by \( y = \frac{X + {i\Delta }}{s}x - \fr... | Yes |
Theorem 19. Let \( {V}_{1}\cdots {V}_{n} \) be a Simson polygon (not necessarily equidistant) with Simson point \( S \) . Let \( i, j, k \in \{ 1,2,\ldots, n\} \), be distinct. Then the circumcircle of the triangle \( T \) formed from lines \( {V}_{i}{V}_{i + 1},{V}_{j}{V}_{j + 1} \) and \( {V}_{k}{V}_{k + 1} \) passes... | Proof. Since the projections of \( S \) into \( {V}_{i}{V}_{i + 1},{V}_{j}{V}_{j + 1} \) and \( {V}_{k}{V}_{k + 1} \) are collinear, \( S \) is a Simson point of the triangle \( T \) . Therefore by the Simson-Wallace Theorem (Theorem 1), \( S \) lies on the circumcircle of \( T \) . | Yes |
Corollary 20. (Lambert's Theorem). The focus of a parabola lies on the circumcircle of a triangle formed by any three tangents to the parabola. | Proof. Taking the limit of a sequence of equidistant Simson polygons gives Lambert’s Theorem for a parabola, since the lines \( {V}_{i}{V}_{i + 1},{V}_{j}{V}_{j + 1},{V}_{k}{V}_{k + 1} \) become tangents in the limit. | No |
Proposition 1. \( {O}_{a} \) and \( {A}^{\prime } \) are the isogonal conjugates in triangle \( {ABC} \) . | Proof. Clearly the lines \( A{O}_{a} \) and \( {AH} \) are isogonal with respect to angle \( A \), since \( O \) and \( H \) are isogonal conjugates. Also,\n\n\[ \angle {A}^{\prime }B{C}_{a} = 2\angle {A}^{\prime }A{C}_{a} = 2\angle {HAB} = 2\angle {OAC} = \angle {O}^{a}{OC} = \angle {O}^{a}{BC}. \]\n\nTherefore, the l... | Yes |
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